Simplify; express your answer in exponential form. Assume $y\neq 0, q\neq 0$. $\dfrac{{(y)^{3}}}{{(y^{2}q^{-1})^{2}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y}$ to the exponent ${3}$ . Now ${1 \times 3 = 3}$ , so ${(y)^{3} = y^{3}}$ In the denominator, we can use the distributive property of exponents. ${(y^{2}q^{-1})^{2} = (y^{2})^{2}(q^{-1})^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y)^{3}}}{{(y^{2}q^{-1})^{2}}} = \dfrac{{y^{3}}}{{y^{4}q^{-2}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{3}}}{{y^{4}q^{-2}}} = \dfrac{{y^{3}}}{{y^{4}}} \cdot \dfrac{{1}}{{q^{-2}}} = y^{{3} - {4}} \cdot q^{- {(-2)}} = y^{-1}q^{2}$.